The cryoscopic contant of water is 1.86 K kg `mol^(-1)`. A 0.01 molal acetic acid solution produces a depression of `0.0194^(@)C` in the freezing point. The degree of dissociation of acetic acid is :

Options

A. zero
B. 0.043
C. 0.43
D. 1

Answer

To solve the problem, we need to find the degree of dissociation (α) of acetic acid in a 0.01 molal solution based on the given depression in freezing point (ΔTf) and the cryoscopic constant (Kf) of water.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Cryoscopic constant of water (Kf) = 1.86 K kg/mol
- Depression in freezing point (ΔTf) = 0.0194 °C
- Molality (m) of acetic acid solution = 0.01 molal

2. **Use the Formula for Depression in Freezing Point:**
The formula relating the depression in freezing point to the molality and Van't Hoff factor (i) is:
\[ \Delta T_f = K_f \times m \times i \]

3. **Rearrange the Formula to Solve for Van't Hoff Factor (i):**
\[ i = \frac{\Delta T_f}{K_f \times m} \]

4. **Substitute the Values into the Formula:**
\[ i = \frac{0.0194}{1.86 \times 0.01} \]

5. **Calculate the Denominator:**
\[ 1.86 \times 0.01 = 0.0186 \]

6. **Calculate the Van't Hoff Factor (i):**
\[ i = \frac{0.0194}{0.0186} \approx 1.043 \]

7. **Relate the Van't Hoff Factor to Degree of Dissociation (α):**
For acetic acid (which dissociates into acetate ions and hydrogen ions), the relationship is:
\[ i = 1 + \alpha \]

8. **Substitute the Value of i to Find α:**
\[ 1.043 = 1 + \alpha \]
\[ \alpha = 1.043 - 1 = 0.043 \]

9. **Conclusion:**
The degree of dissociation (α) of acetic acid in the solution is:
\[ \alpha = 0.043 \]

### Final Answer:
The degree of dissociation of acetic acid is **0.043**.